July 9, 2020

Exam 4

Physics 207 – Introductory Physics I
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
Fall 2011

Exam 4
Circular Motion, Gravitation, Work & Energy

§ Problem 1

Two stars of equal mass are gravitationally bound in a binary star system. When this happens they orbit the point which is midway between them. As shown in the picture, the two stars, due to their mutual gravitational attraction, will continue to move in the circular orbit shown. Let

\displaystyle m \displaystyle= \displaystyle 1.99\times 10^{{30}}\mathrm{kg}
\displaystyle r \displaystyle= \displaystyle 149.6\times 10^{9}\mathrm{m}

The stars are the same size as our Sun, and the orbital radius is the same size as the mean Earth–Sun distance, but that’s just a fun coincidence.

§ (a)

What is the strength of the gravitational force felt by one star due to the other. Clearly indicate the equation used, then substitute the numbers, and finally present the numerical answer. (10 points)

Note that the distance between the two stars is twice the radius of the circle.

\displaystyle F \displaystyle= \displaystyle G\frac{m_{1}m_{2}}{r_{{m_{1}m_{2}}}^{2}}
\displaystyle= \displaystyle G\frac{mm}{(2r)^{2}}
\displaystyle= \displaystyle\left(6.67\times 10^{{-11}}\frac{\mathrm{Nm}^{2}}{\mathrm{kg}^{2}}\right)\frac{1}{4}\left(\frac{1.99\times 10^{{30}}\mathrm{kg}}{149.6\times 10^{9}\mathrm{m}}\right)^{2}
\displaystyle= \displaystyle 2.95\times 10^{{27}}\mathrm{N}

§ (b)

What will be the speed v of either star. (It will be the same for both stars, so you only need to solve once.) (10 points)

We need to equate the force from part (a) to supply the centripetal force to keep the star moving on a circle of radius r. I choose to go back to the expression for the force rather than just plugging in the number, but either approach would be okay.

\displaystyle F \displaystyle= \displaystyle\frac{mv^{2}}{r}
\displaystyle G\frac{m^{2}}{4r^{2}} \displaystyle= \displaystyle\frac{mv^{2}}{r}
\displaystyle G\frac{m}{4r} \displaystyle= \displaystyle v^{2}
\displaystyle v \displaystyle= \displaystyle\sqrt{G\frac{m}{4r}}
\displaystyle= \displaystyle\sqrt{\frac{1}{4}\left(6.67\times 10^{{-11}}\frac{\mathrm{Nm}^{2}}{\mathrm{kg}^{2}}\right)\frac{1.99\times 10^{{30}}\mathrm{kg}}{149.6\times 10^{9}\mathrm{m}}}
\displaystyle= \displaystyle 1.49\times 10^{4}\frac{\mathrm{m}}{\mathrm{s}}

§ (c)

What is the angular velocity \omega of either star? (5 points)

\displaystyle\omega \displaystyle= \displaystyle\frac{v}{r}
\displaystyle= \displaystyle\frac{1.49\times 10^{4}\frac{\mathrm{m}}{\mathrm{s}}}{149.6\times 10^{9}\mathrm{m}}
\displaystyle= \displaystyle 9.96\times 10^{{-8}}\frac{\mathrm{rad}}{\mathrm{s}}

§ (d)

What is the orbital period T of either star? (The period is the time required to orbit once.) (5 points)

One orbit is 2\pi\;\mathrm{rad} and takes place in time t=T.

\displaystyle\theta \displaystyle= \displaystyle\omega t
\displaystyle 2\pi \displaystyle= \displaystyle\omega T
\displaystyle T \displaystyle= \displaystyle\frac{2\pi}{\omega}
\displaystyle= \displaystyle\frac{2\pi}{9.96\times 10^{{-8}}\frac{\mathrm{rad}}{\mathrm{s}}}
\displaystyle= \displaystyle 6.31\times 10^{7}\mathrm{s}\times\frac{1\;\mathrm{h}}{3600\;\mathrm{s}}\times\frac{1\;\mathrm{day}}{24\;\mathrm{h}}\times\frac{1\;\mathrm{year}}{365.25\;\mathrm{day}}
\displaystyle= \displaystyle 2.00\;\mathrm{Earth\; years}

§ Problem 2

On a theme-park thrill ride a roller coaster travels over the outside of a circular loop of radius r=5.0\;\mathrm{m} with a speed of v=9.9\;\mathrm{m}/\mathrm{s}. A safety harness is used to hold each rider safely in the car. What force must the safety harness provide when the rider, mass m=65\;\mathrm{kg}, is exactly at the top of the loop? (20 points)

The normal force, the weight and the centripetal acceleration all point from the top of the circle towards the center of the circle. This is virtually identical to problems in the homework.

\displaystyle F_{N}+mg \displaystyle= \displaystyle\frac{mv^{2}}{r}
\displaystyle F_{N} \displaystyle= \displaystyle\frac{mv^{2}}{r}-mg
\displaystyle= \displaystyle\frac{\left(65\;\mathrm{kg}\right)\left(9.9\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}}{5.0\;\mathrm{m}}-\left(65\;\mathrm{kg}\right)\left(9.80\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)
\displaystyle= \displaystyle 640\;\mathrm{N}


\displaystyle W \displaystyle= \displaystyle mg
\displaystyle= \displaystyle\left(65\;\mathrm{kg}\right)\left(9.80\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)
\displaystyle= \displaystyle 640\;\mathrm{N}

The numbers were chosen so that the force of the safety harness is equal to the person’s weight.

§ Problem 3

A regulation ping-pong ball with a mass of m=2.7\;\mathrm{g} is shown in the figure. It is at rest on a massless compressed coil spring with a spring constant k=3.2\;\mathrm{N}/\mathrm{m}. The spring as it is shown is compressed by a distance of 5.0\;\mathrm{cm}.
In the original configuration shown:

§ (a)

What is the initial elastic potential energy stored in the spring? (4 points)

\displaystyle\mathrm{PE}_{{\mathrm{elastic},i}} \displaystyle= \displaystyle\frac{1}{2}kx^{2}
\displaystyle= \displaystyle\frac{1}{2}\left(3.2\frac{\mathrm{N}}{\mathrm{m}}\right)\left(-0.050\;\mathrm{m}\right)^{2}
\displaystyle= \displaystyle 4.0\times 10^{{-3}}\mathrm{J}

§ (b)

What is the initial gravitational potential energy of the ping-pong ball? (4 points)

\displaystyle\mathrm{PE}_{{\mathrm{gravity},i}} \displaystyle= \displaystyle mgh
\displaystyle= \displaystyle mg0
\displaystyle= \displaystyle 0\;\mathrm{J}

§ (c)

What is the initial kinetic energy of the ping-pong ball? (4 points)

\displaystyle\mathrm{KE}_{i} \displaystyle= \displaystyle\frac{1}{2}mv^{2}
\displaystyle= \displaystyle\frac{1}{2}m0^{2}
\displaystyle= \displaystyle 0\;\mathrm{J}

The spring is allowed to expand pushing the ball upwards. When the spring has expanded to its equilibrium position the ball will separate from the spring.

§ (d)

What is the elastic potential energy stored in the spring at the instant the ball separates? (4 points)

\displaystyle\mathrm{PE}_{{\mathrm{elastic},2}} \displaystyle= \displaystyle\frac{1}{2}k0^{2}
\displaystyle= \displaystyle 0\;\mathrm{J}

§ (e)

What is the gravitational potential energy of the ping-pong ball at the instant it separates from the spring? (4 points)

\displaystyle\mathrm{PE}_{{\mathrm{gravity},2}} \displaystyle= \displaystyle mgh
\displaystyle= \displaystyle\left(0.0027\;\mathrm{kg}\right)\left(9.80\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)\left(0.050\;\mathrm{m}\right)
\displaystyle= \displaystyle 1.32\times 10^{{-3}}\mathrm{J}

§ (f)

What is the kinetic energy of the ping-pong ball at the instant it separates from the spring? (4 points)

First we should take note of the total energy in the system which we may compute from the original configuration.

\displaystyle E \displaystyle= \displaystyle\mathrm{PE}_{{\mathrm{elastic},i}}+\mathrm{PE}_{{\mathrm{gravity},i}}+\mathrm{KE}_{i}
\displaystyle= \displaystyle 4.0\times 10^{{-3}}\mathrm{J}+0\;\mathrm{J}+0\;\mathrm{J}
\displaystyle= \displaystyle 4.0\times 10^{{-3}}\mathrm{J}

At the instant of separation,

\displaystyle E \displaystyle= \displaystyle\mathrm{PE}_{{\mathrm{elastic},2}}+\mathrm{PE}_{{\mathrm{gravity},2}}+\mathrm{KE}_{2}
\displaystyle 4.0\times 10^{{-3}}\mathrm{J} \displaystyle= \displaystyle 0\;\mathrm{J}+1.32\times 10^{{-3}}\mathrm{J}+\frac{1}{2}mv^{2}
\displaystyle\frac{1}{2}mv^{2} \displaystyle= \displaystyle 4.0\times 10^{{-3}}\mathrm{J}-1.32\times 10^{{-3}}\mathrm{J}
\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{2(2.68\times 10^{{-3}}\mathrm{J})}{0.0027\;\mathrm{kg}}}
\displaystyle= \displaystyle 1.41\frac{\mathrm{m}}{\mathrm{s}}

§ (g)

What is the maximum height h reached by the ping-pong ball? (4 points)

Using conservation of energy

\displaystyle E \displaystyle= \displaystyle\mathrm{PE}_{{\mathrm{elastic},f}}+\mathrm{PE}_{{\mathrm{gravity},f}}+\mathrm{KE}_{f}
\displaystyle 4.0\times 10^{{-3}}\mathrm{J} \displaystyle= \displaystyle 0\;\mathrm{J}+0\;\mathrm{J}+mgh
\displaystyle h \displaystyle= \displaystyle\frac{4.0\times 10^{{-3}}\mathrm{J}}{mg}
\displaystyle= \displaystyle\frac{4.0\times 10^{{-3}}\mathrm{J}}{(0.0027\;\mathrm{kg})\left(9.80\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)}
\displaystyle= \displaystyle 0.15\;\mathrm{m}
\displaystyle= \displaystyle 15\;\mathrm{cm}

In this last step one could also use kinematic equations since there was not a requirement to use conservation of energy.

§ Problem 4

A skier approaches a ski jump which makes an angle of \theta=30^{0} with the horizontal. The jump will lift the skier a height h=2.0\;\mathrm{m}. The skier has a speed of v_{i}=10.0\;\mathrm{m}/\mathrm{s} when she reaches the start of the jump.

§ (a)

Use conservation of energy to find the speed v_{f} as the skier leaves the jump assuming that friction is negligible. (9 points)

Using conservation of energy

\displaystyle\mathrm{PE}_{i}+\mathrm{KE}_{i} \displaystyle= \displaystyle\mathrm{PE}_{f}+\mathrm{KE}_{f}
\displaystyle mgh_{i}+\frac{1}{2}mv_{i}^{2} \displaystyle= \displaystyle mgh_{f}+\frac{1}{2}mv_{f}^{2}
\displaystyle gh_{i}+\frac{1}{2}v_{i}^{2} \displaystyle= \displaystyle gh_{f}+\frac{1}{2}v_{f}^{2}
\displaystyle 0+\frac{1}{2}v_{i}^{2} \displaystyle= \displaystyle gh_{f}+\frac{1}{2}v_{f}^{2}
\displaystyle v_{f} \displaystyle= \displaystyle\sqrt{v_{i}^{2}-2gh_{f}}
\displaystyle= \displaystyle\sqrt{\left(10.0\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}-2\left(9.80\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)(2.0\;\mathrm{m})}
\displaystyle= \displaystyle 7.8\frac{\mathrm{m}}{\mathrm{s}}

§ (b)

Use conservation of energy to find the speed v_{f} as the skier leaves the jump if the coefficient of kinetic friction is \mu _{k}=0.20. (9 points)

This problem differs from part (a) in that some energy will be dissipated through friction. If W is the magnitude of the work done by friction, that is, taken as a positive number, then we may modify the energy-conservation equation from part (a) as

\displaystyle\frac{1}{2}mv_{i}^{2}-W \displaystyle= \displaystyle mgh_{f}+\frac{1}{2}mv_{f}^{2}

This means that v_{f} should come out smaller. To compute the work,

\displaystyle W \displaystyle= \displaystyle F_{{\mathrm{fr}}}d

The distance d is the hypotenuse of the triangle shown, so

\displaystyle\sin 30^{0} \displaystyle= \displaystyle\frac{h_{f}}{d}
\displaystyle d \displaystyle= \displaystyle 2h_{f}

Where I have used \sin 30^{0}=1/2.

\displaystyle F_{{\mathrm{fr}}} \displaystyle= \displaystyle\mu _{k}F_{N}
\displaystyle= \displaystyle\mu _{k}mg\cos 30^{0}
\displaystyle= \displaystyle\mu _{k}mg\frac{\sqrt{3}}{2}

Ignoring the sign from the \cos 180^{0} since we want the magnitude of the work,

\displaystyle W \displaystyle= \displaystyle F_{{\mathrm{fr}}}d
\displaystyle= \displaystyle\mu _{k}mg\frac{\sqrt{3}}{2}(2h_{f})
\displaystyle= \displaystyle\mu _{k}mg\sqrt{3}h_{f}

We now substitute that into the energy-conservation equation.

\displaystyle\frac{1}{2}mv_{i}^{2}-\mu _{k}mg\sqrt{3}h_{f} \displaystyle= \displaystyle mgh_{f}+\frac{1}{2}mv_{f}^{2}


\displaystyle v_{f} \displaystyle= \displaystyle\sqrt{v_{i}^{2}-gh_{f}(2+\sqrt{3}\mu _{k})}
\displaystyle= \displaystyle\sqrt{\left(10.0\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}-\left(9.80\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)(2.0\;\mathrm{m})\left(2+\sqrt{3}(0.20)\right)}
\displaystyle= \displaystyle 7.3\frac{\mathrm{m}}{\mathrm{s}}

which is a bit less than the result found in part (a), as expected.


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